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(B) Let the intensities of the two waves be $I_1$ and $I_2$.Given that the ratio of intensities is $I_1 : I_2 = 9 : 1$. Let $I_1 = 9x$ and $I_2 = x$.T

Vedclass · Accepted Answer

$4:1$

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(B) Let the intensities of the two waves be $I_1$ and $I_2$.Given that the ratio of intensities is $I_1 : I_2 = 9 : 1$. Let $I_1 = 9x$ and $I_2 = x$.The formula for the ratio of maximum to minimum intensity in an interference pattern is given by $\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2$.Substituting the values,we get $\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{9x} + \sqrt{x}}{\sqrt{9x} - \sqrt{x}} \right)^2$.This simplifies to $\frac{I_{\max}}{I_{\min}} = \left( \frac{3\sqrt{x} + \sqrt{x}}{3\sqrt{x} - \sqrt{x}} \right)^2 = \left( \frac{4\sqrt{x}}{2\sqrt{x}} \right)^2$.$\frac{I_{\max}}{I_{\min}} = (2)^2 = 4$.Thus,the ratio of maximum to minimum intensity is $4 : 1$.

Two waves having the intensities in the ratio of $9 : 1$ produce interference. The ratio of maximum to the minimum intensity is equal to

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